Problem: You have found the following ages (in years) of all $6$ lions at your local zoo: $ 13,\enspace 2,\enspace 1,\enspace 5,\enspace 2,\enspace 7$ What is the average age of the lions at your zoo? What is the standard deviation? Round your answers to the nearest tenth. Average age: $ $
Explanation: Because we have data for all $6$ lions at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$. To find the population mean, add up the values of all $6$ ages and divide by $6$. $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{13 + 2 + 1 + 5 + 2 + 7}{{6}} = {5\text{ years old}} $ Find the squared deviations from the mean for each lion. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $13$ years $8$ years $64$ years $^2$ $2$ years $-3$ years $9$ years $^2$ $1$ year $-4$ years $16$ years $^2$ $5$ years $0$ years $0$ years $^2$ $2$ years $-3$ years $9$ years $^2$ $7$ years $2$ years $4$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean, we can find the variance $({\sigma^2})$, without introducing any bias, by simply averaging the squared deviations from the mean: $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{64} + {9} + {16} + {0} + {9} + {4}} {{6}} $ $ {\sigma^2} = \dfrac{{102}}{{6}} = {17\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$. ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{17\text{ years}^2}} = {4.1\text{ years}} $ The average lion at the zoo is $5$ years old. There is a standard deviation of $4.1$ years.